Question

Find the derivative of $sin\text{ }x$ with respect to x from first principles.

Answer 1

Hint: The derivative using the first principal is given by the formula $\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right]$, where f(x) is the function to be differentiation with respect to x, and h is tending to zero.
Complete step-by-step answer:
In the question, we have to find the derivative of $sin\text{ }x$ with respect to x from first principles.
So, for that we will use the formula $\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right]$, here the function to be differentiated is $f\left( x \right)=\text{ }sin\text{ }x$. Now as per the formula, we have the required derivative as:

& \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right] \\\ & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\sin \left( x+h \right)-\sin \left( x \right)}{h} \right) \\\ \end{aligned}$$ Here we can see that the limit is of the form $$\dfrac{0}{0}$$, as we have $$\underset{h\to \;0}{\mathop{\lim }}\,\,\,\sin \left( x+h \right)=\sin \left( x \right)$$ So $$\underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\sin \left( x+h \right)-\sin \left( x \right)}{h} \right)=\dfrac{0}{0}$$ Hence, here we will use the L-Hopitals rule, where we differentiate the numerator and the denominator separately, as follows: $$\begin{aligned} & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\sin \left( x+h \right)-\sin \left( x \right)}{h} \right) \\\ & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\dfrac{d}{dh}\left( \sin \left( x+h \right)-\sin \left( x \right) \right)}{\dfrac{d}{dh}\left( h \right)} \right) \\\ & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\left( \cos \left( x+h \right)-0 \right)}{\left( 1 \right)} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \left( \dfrac{d\left( \sin \left( x+h \right) \right)}{dh}=\cos \left( x+h \right),\,\,\,\dfrac{d\left( \sin \left( x \right) \right)}{dh}=0 \right) \\\ & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \cos \left( x+h \right) \right) \\\ & \Rightarrow \cos \left( x \right) \\\ \end{aligned}$$ So final, we can say that $$\begin{aligned} & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right] \\\ & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\sin \left( x+h \right)-\sin \left( x \right)}{h} \right) \\\ & \Rightarrow \cos \left( x \right) \\\ \end{aligned}$$ Hence the derivative of $$\sin \left( x \right)$$ is $$\cos \left( x \right)$$. Note: The limit of the expression is to be found carefully, and when applying the L-Hopitals’ rule, we will separately differentiate the numerator and the denominator and will not use the quotient rule of differentiation.